What are the roots of #2x-5=-3x^2#?

Answer 1

Isaiah Anthony

#x={-5/3,1}#

#2x-5=-3x^2#

#3x^2+2x-5=0#

#(x-1)(3x+5)=0#

#x-1=0" "rArr" "x=1#

#3x+5=0" "rArr" "x=-5/3#

#x={-5/3,1}#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eva Adamson

To find the roots of the equation 2x - 5 = -3x^2, we first rearrange it into a quadratic equation form (ax^2 + bx + c = 0), where a, b, and c are coefficients.

Given equation: 2x - 5 = -3x^2

Rearranging: 3x^2 + 2x - 5 = 0

Now, we can use the quadratic formula to find the roots: x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 3, b = 2, and c = -5.

Plugging in the values: x = [-(2) ± √((2)^2 - 4(3)(-5))] / (2 * 3)

Calculating under the square root: √(4 + 60) √(64) = 8

Now, substituting back into the equation: x = [-(2) ± 8] / 6

This gives two possible roots: x1 = (-2 + 8) / 6 = 6 / 6 = 1 x2 = (-2 - 8) / 6 = -10 / 6 = -5/3

So, the roots of the equation 2x - 5 = -3x^2 are x = 1 and x = -5/3.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.